Question

Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x − 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.

Answer 1

The equation of the straight line passing through the point of intersection of x + y = 4 and 2x − 3y = 1 is

x + y − 4 + λ(2x − 3y − 1) = 0

$\Rightarrow$(1 + 2λ)x + (1 − 3λ)y − 4 − λ = 0 ... (1)

$\Rightarrow y=-\left(\frac{1+2\lambda }{1-3\lambda }\right)x+\frac{4+\lambda }{1-3\lambda }$

The equation of the line with intercepts 5 and 6 on the axis is

$\frac{x}{5}+\frac{y}{6}=1$ ... (2)

The slope of this line is $-\frac{6}{5}$.

The lines (1) and (2) are perpendicular.

$$\begin{gathered} \therefore -\frac{6}{5}\times \left(-\frac{1+2\lambda }{1-3\lambda }\right)=-1 \\ \Rightarrow \lambda =\frac{11}{3} \end{gathered}$$

Substituting the values of λ in (1), we get the equation of the required line.

$$\begin{gathered} \Rightarrow \left(1+\frac{22}{3}\right)x+\left(1-11\right)y-4-\frac{11}{3}=0 \\ \Rightarrow 25x-30y-23=0 \end{gathered}$$